Contents

# Contents

## Definition

### Implicit definition

In real analysis, the reciprocal $\frac{1}{x}$ is a partial function implicitly defined over the non-zero real numbers by the equation $x \left(\frac{1}{x}\right) = 1$. This is the definition commonly used when defining the real numbers as a field.

### By the exponential and natural logarithm functions

The reciprocal $\frac{1}{(-)}:(-\infty,0)\union(0,\infty)\to\mathbb{R}$ is piecewise defined as

$\frac{1}{x} \coloneqq \begin{cases} -e^{-\ln(-x)} & x \in (-\infty,0) \\ e^{-\ln(x)} & x \in (0,\infty) \end{cases}$

This definition implies that the reciprocal is analytic in each of the two connected components of the domain.

### By power series

Let us define the functions $f \colon (-1,1)\to\mathbb{R}$ and $g:(-1,1)\to\mathbb{R}$ from the open subinterval of the real numbers $(-1,1) \subset \mathbb{R}$ to the real numbers $\mathbb{R}$ as the locally convergent power series

$f(x)\coloneqq \sum_{n=0}^{\infty} x^n$
$g(x)\coloneqq \sum_{n=0}^{\infty} (-1)^n x^n$

The reciprocal $\frac{1}{(-)}:(-\infty,0)\union(0,\infty)\to\mathbb{R}$ is then piecewise defined as

$\frac{1}{x} \coloneqq \begin{cases} \lim_{a\to 0^-} (-a) \sum_{n=0}^{\infty} a^n (x-f(a+1))^n & x \in (-\infty,0) \\ \lim_{a\to 0^+} a \sum_{n=0}^{\infty} (-a)^n (x+g(a-1))^n & x \in (0,\infty) \end{cases}$

Equivalently, given an element $p \in (0, 1)$, the reciprocal $\frac{1}{(-)}:(-\infty,0)\union(0,\infty)\to\mathbb{R}$ is then piecewise defined as

$\frac{1}{x} \coloneqq \begin{cases} \lim_{i \to \infty} p^i \sum_{n=0}^{\infty} (-p^i)^n (x-f(-p^i+1))^n & x \in (-\infty,0) \\ \lim_{i \to \infty} p^i \sum_{n=0}^{\infty} (-p^i)^n (x+g(p^i-1))^n & x \in (0,\infty) \end{cases}$

This definition implies that the reciprocal is analytic in each of the two connected components of the domain.

Last revised on December 4, 2022 at 21:58:18. See the history of this page for a list of all contributions to it.